Indeterminate form 0/0

$$\displaystyle\lim_{x \to 2}{\dfrac{x^2-3x+2}{x-2}}=\dfrac{4-6+2}{0}=\dfrac{0}{0}$$

Since $2$ cancels the polynomial of the numerator, we factor it:

$$x^2-3x+2=(x-1)\cdot(x-2)$$

$$\displaystyle\lim_{x \to 2}{\dfrac{(x-1)(x-2)}{(x-2)}}=\lim_{x \to 2}{(x-1)}=2-1=1$$

$$\displaystyle\lim_{x \to 3}{\dfrac{x-3}{1-\sqrt{x-2}}}=\dfrac{0}{0}$$

Let's multiply and divide by the conjugate:

$$\displaystyle\lim_{x \to 3}{\dfrac{(x-3)(1+\sqrt{x-2})}{(1-\sqrt{x-2})(1+\sqrt{x-2})}}=\lim_{x \to 3}{\dfrac{(x-3)(1+\sqrt{x-2})}{1^2-(\sqrt{x-2})^2}}=$$

$$=\displaystyle\lim_{x \to 3}{\dfrac{(x-3)(1+\sqrt{x-2})}{1-x+2}}=\lim_{x \to 3}{\dfrac{(x-3)(1+\sqrt{x-2})}{-(x-3)}}=$$

$$=\displaystyle\lim_{x \to 3}{-(1+\sqrt{x-2})}=-(1+\sqrt{3-2})=-(1+1)=-2$$